Answer:
2.34 mins
Explanation:
The regression equation in mathematical terms, shows the relationships between a dependent variable (reaction time) and one or more independent variables (exposure to a chemical level).
It also shows the effect of exposure to chemical on the reaction time.
The linear regression equation Y=2.3X-1.8 can be used to make predictions about about Y which is the reaction time for a given value of X which is the level of exposure to chemical.
From the question, X (level of exposure to chemical) is 1.8
The regression equation becomes
Y=2.3(1.8) - 1.8
Y= 4.4 - 1.8
Y= 2.34
There fore, the reaction time is 2.34 mins
Answer: An existing theory is modified so that it can explain both the old and new observations.
Explanation:
To find the number of moles from a mass given, simply look to the formula n (moles) = m (mass, g) / MM (molar mass).
Mass was given, 36.04
Molar mass is the total atomic mass of all the atoms present. Water is H20, so that means 2 hydrogen and 1 oxygen. The atomic mass of hydrogen is 1 and atomic mass of oxygen is 16. Therefore MM= 1 + 1 + 16= 18.
Plug that value in and the full equation is
n = 36.04/18
n = 2.002 moles
= 2 moles
Answer:
2Cr³⁺(aq) + 3S²⁻(aq) ----> Cr₂S₃(s)
Explanation:
When aqueous solutions of chromium(III) chloride and ammonium sulfide are mixed together, chromium (III) chloride and ammonium sulfide undergoes a double displacement reaction to produce chromium (iii) sulphide as a precipitate and ammonium chloride which remains in solution.
The general equation of the reaction is given below:
2CrCl₃(aq) + 3(NH₄)₂S(aq) ----> Cr₂S₃(s) + 6NH₄Cl(aq)
The net equation of the reaction is given below:
2Cr³⁺(aq) + 3S²⁻(aq) ----> Cr₂S₃(s)
Answer:
6 days
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Half life (t½) =?
Next, we shall determine the decay constant. This can be obtained as follow:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Decay constant (K) =?
Log (N₀/N) = kt / 2.303
Log (100/6.25) = k × 24 / 2.303
Log 16 = k × 24 / 2.303
1.2041 = k × 24 / 2.303
Cross multiply
k × 24 = 1.2041 × 2.303
Divide both side by 24
K = (1.2041 × 2.303) / 24
K = 0.1155 /day
Finally, we shall determine the half-life of the isotope as follow:
Decay constant (K) = 0.1155 /day
Half life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 0.1155
t½ = 6 days
Therefore, the half-life of the isotope is 6 days